Simplify the following

(i) $\sqrt{45}-3 \sqrt{20}+4 \sqrt{5}=\sqrt{3 \times 3 \times 5}-3 \sqrt{2 \times 2 \times 2}+4 \sqrt{5}$

$=3 \sqrt{5}-3 \times 2 \sqrt{5}+4 \sqrt{5}$

$=3 \sqrt{5}-6 \sqrt{5}+4 \sqrt{5}=\sqrt{5}$

(ii) $\frac{\sqrt{24}}{8}+\frac{\sqrt{54}}{9}=\frac{\sqrt{2 \times 2 \times 2 \times 3}}{8}+\frac{\sqrt{3 \times 3 \times 3 \times 2}}{9}$

$=\frac{2 \sqrt{6}}{8}+\frac{3 \sqrt{6}}{9}=\frac{\sqrt{6}}{4}+\frac{\sqrt{6}}{3}$

$=\frac{3 \sqrt{6}+4 \sqrt{6}}{12}=\frac{7 \sqrt{6}}{12}$

(iii) $\sqrt[4]{12} \times \sqrt[7]{6}=(12)^{1 / 4} \times(6)^{1 / 7}$     $\left[\because \sqrt[n]{a}=a^{1 / n}\right]$

$=(2 \times 2 \times 3)^{1 / 4} \times(2 \times 3)^{1 / 7}=2^{1 / 4} \cdot 2^{1 / 4} \cdot 3^{1 / 4} \cdot 2^{1 / 7} \cdot 3^{1 / 7}$

$=2^{\frac{1}{4}+\frac{1}{4}+\frac{1}{7}} \times 3^{\frac{1}{4}+\frac{1}{7}}$                    $\left[\because x^{m} \cdot x^{n}=x^{m+n}\right]$

 $=2^{9 / 14} \times 3^{11 / 28}=\sqrt[14]{2^{9}} \sqrt[28]{3^{11}}$             $\left[\because \sqrt[n]{a}=\sqrt[m n]{a^{m}}\right]$

(iv) $4 \sqrt{28}+3 \sqrt{7}+\sqrt[3]{7}=(4 \sqrt{4 \times 7}+3 \sqrt{7})+\sqrt[3]{7}$

$=\left(\frac{8 \sqrt{7}}{3 \sqrt{7}}\right)+(7)^{\frac{1}{3}}$                      $\left[\because \sqrt[n]{a}=a^{1 / n}\right]$

$=\frac{8}{3}+7^{\frac{1}{3}}=\frac{8}{3 \sqrt[3]{7}}$

(v) $3 \sqrt{3}+2 \sqrt{27}+\frac{7}{\sqrt{3}}=3 \sqrt{3}+2 \sqrt{3 \times 3 \times 3}+\frac{7}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$

$=3 \sqrt{3}+6 \sqrt{3}+\frac{7 \sqrt{3}}{3}=9 \sqrt{3}+\frac{7 \sqrt{3}}{3}$

$=\frac{27 \sqrt{3}+7 \sqrt{3}}{3}=\frac{34 \sqrt{3}}{3}$

(vi) $(\sqrt{3}-\sqrt{2})^{2}=(\sqrt{3})^{2}+(\sqrt{2})^{2}-2 \sqrt{3} \times \sqrt{2}$               [using identity, $(a-b)^{2}=a^{2}+b^{2}-2 a b$ ]

$=3+2-2 \sqrt{3 \times 2}=5-2 \sqrt{6}$

(vii) $\sqrt[4]{81}-8 \sqrt[3]{216}+15 \sqrt[5]{32}+\sqrt{225}$

$=(81)^{\frac{1}{4}}-8 \times(216)^{\frac{1}{3}}+15 \times(32)^{\frac{1}{5}}+\sqrt{(15)^{2}}$           $\left[\because \sqrt[m]{a}=a^{1 / m}\right]$

$=\left(3^{4}\right)^{\frac{1}{4}}-8 \times\left(6^{3}\right)^{\frac{1}{3}}+15 \times\left(2^{5}\right)^{\frac{1}{5}}+15$

$=3^{4 \times \frac{1}{4}}-8 \times 6^{3 \times \frac{1}{3}}+15 \times 2^{5 \times \frac{1}{5}}+15$             $\left[\because\left(a^{m}\right)^{n}=a^{m n}\right]$

$=3^{1}-8 \times 6^{1}+15 \times 2^{1}+15$

$=3-48+30+15$

$=48-48=0$

(viii) $\frac{3}{\sqrt{8}}+\frac{1}{\sqrt{2}}=\frac{3}{\sqrt{2 \times 2 \times 2}}+\frac{1}{\sqrt{2}}=\frac{3}{2 \sqrt{2}}+\frac{1}{\sqrt{2}}$

$=\frac{3+2}{2 \sqrt{2}}=\frac{5}{2 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$               [multiplying numerator and denominator by $\sqrt{2}$ ]

$=\frac{5 \sqrt{2}}{2 \times 2}=\frac{5 \sqrt{2}}{4}$

(ix) $\frac{2 \sqrt{3}}{3}-\frac{\sqrt{3}}{6}=\frac{4 \sqrt{3}-\sqrt{3}}{6}=\frac{3 \sqrt{3}}{6}=\frac{\sqrt{3}}{2}$